Sand Pile Volume

Jason

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A sand pile has a base that covers a region in the $$\displaystyle xy$$ pane that is bounded by the line $$\displaystyle y = 2x$$ and the parabola $$\displaystyle x^{2} + y = 3$$. The sand's height above the point $$\displaystyle (x,y)$$ is $$\displaystyle x^{2}$$. Express the volume of sand as a double and triple integral and, next, calculate the volume.

MarkFL

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The first thing I would do here is plt the bounded region:

We see this is a type I region, so for a double iterated integral, we will want to integrate with respect to $$y$$ first (inner) and then with respect to $$x$$ (outer). To determine the outer limits, we can equate the two functions making up the region boundaries:

$$\displaystyle 2x=3-x^2$$

$$\displaystyle x^2+2x-3=0$$

$$\displaystyle (x+3)(x-1)=0$$

$$\displaystyle x\in\{-3,1\}$$

And so:

$$\displaystyle V=\int_{-3}^{1} x^2\int_{2x}^{3-x^2}\,dy\,dx$$

Can you proceed?

anemone and Jason

Jason

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How were the boundaries gotten for the other integral - the one without $$\displaystyle 1$$ and $$\displaystyle -3$$ limits of integration?

OH wait (post edited). I see now! But how did you determine which limit of integration was on the bottom vs which one is on the top?

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MarkFL

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The line is below the parabola (which we can see from the plot, or we could show with an inequality), so the line will be the bottom limit while the parabola will serve as the upper limit.

anemone

Jason

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$$\displaystyle \int_{-3}^{1}\,x^{2} \,dy$$

$$\displaystyle \dfrac{x^{3}}{3}$$ - evaluated at $$\displaystyle -3$$ and $$\displaystyle 1$$

$$\displaystyle [\dfrac{(1)^{3}}{3}] - [\dfrac{(-3)^{3}}{3}] = \dfrac{28}{3}$$

Cannot be done further as a double integral. There aren't any variables left for the 2nd part.

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MarkFL

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What you want to do is evaluate the inner integral first:

$$\displaystyle V=\int_{-3}^{1} x^2\int_{2x}^{3-x^2}\,dy\,dx=\int_{-3}^{1} x^2\left((3-x^2)-(2x)\right)\,dx=\int_{-3}^{1} 3x^2-2x^3-x^4\,dx$$

Now, you can evaluate the outer integral.

anemone

MarkFL

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As a triple integral, I would write:

$$\displaystyle V=\int_{-3}^{1} \int_{2x}^{3-x^2} \int_0^{x^2}\,dz\,dy\,dx$$

anemone

MarkFL

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Suppose in the double integral, I wished to reverse the order of integration...what issue would I face?

anemone

Jason

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Suppose in the double integral, I wished to reverse the order of integration...what issue would I face?
It would be the one I came up with where - once you get the next integral, there's nothing to work with. (guess)

MarkFL

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It would be the one I came up with where - once you get the next integral, there's nothing to work with. (guess)
If we reverse the order of integration in the double integral I gave, we would be treating the bounded area as type II, which means we are using horizontal rather than vertical strips. Using horizontal strips, and looking at the bounded region, do you see that the boundary on the right changes from one function to another?

Jason

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If we reverse the order of integration in the double integral I gave, we would be treating the bounded area as type II, which means we are using horizontal rather than vertical strips. Using horizontal strips, and looking at the bounded region, do you see that the boundary on the right changes from one function to another?
So we would certainly need to know what order of integration to use before we attempt the problem. How could we find out?

MarkFL

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So we would certainly need to know what order of integration to use before we attempt the problem. How could we find out?
We can evaluate the integral in either order, but I chose the order I did for simplicity. If we treat the area as type II, then we will need two separate integrals. Sometime though, we find that an multiple iterated integral evaluated in one order results in an anti-derivative that is not expressible in elementary functions, but if we reverse or change the order of integration, we don't run into that as an issue. So, it becomes useful to both look at which type is going to be easier, and to be able to switch the order in cases where we need to to be able to actually evaluate the integral in elementary terms.

That's why I plotted the bounded area first, so that I could see which order made the most sense as a way to begin.

Jason

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We can evaluate the integral in either order, but I chose the order I did for simplicity. If we treat the area as type II, then we will need two separate integrals. Sometime though, we find that an multiple iterated integral evaluated in one order results in an anti-derivative that is not expressible in elementary functions, but if we reverse or change the order of integration, we don't run into that as an issue. So, it becomes useful to both look at which type is going to be easier, and to be able to switch the order in cases where we need to to be able to actually evaluate the integral in elementary terms.

That's why I plotted the bounded area first, so that I could see which order made the most sense as a way to begin.
I don't really get what you mean. But I know doing it the way I tried at first caused it to disappear - making it impossible to solve.

MarkFL

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Look at the difference between using vertical strips going from left to right, where the lower boundary of the strips is the line and the upper boundary is the parabola vs. using horizontal strips going from bottom to top...do you see there is a point (1,2) where the rightmost boundary of the strips changes from being on the line to being on the parabola? This would require us to set up two separate integrals, and that's what I wanted to avoid when I chose the order I did.

Jason

Jason

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Right, well, the fact the integral disappeared going one way - showed that another way would have be used or it would be impossible.

MarkFL

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If set up properly, it would give the same result going either way...one is just simpler than the other. What you posted as your attempt wasn't set up correctly.

Jason

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What mistakes were made in my attempt, aside from not going from the inside out?

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